# Thread: Solving Algebraically - Simultaneous equations.

1. ## Solving Algebraically - Simultaneous equations.

Solve Algebraically - These Simultaneous equations.
y=3x-2
x^2+y^2=20
Here is what I've done so far, please correct me if i'm wrong as always.
x^2+(3x-2)(3x-2)=20 x^2+9x^2-6x-6x-16=0 10x^2-12x-16=0
Can anyone solve this without using the quadratic equation? Many Thanks!

2. Hello,

Correct working

But why don't you want to use the quadratic formula ?

The roots of $\displaystyle ax^2+bx+c=0$ are $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
that's much simpler !

(I tried to do it by completing the square, and that was quite useless)

3. Originally Posted by BabyMilo
Solve Algebraically - These Simultaneous equations. Here is what I've done so far, please correct me if i'm wrong as always. Can anyone solve this without using the quadratic equation? Many Thanks!
It's pretty easy to factor it. You can divide $\displaystyle 10x^2-12x-16=0$ by 2 to get $\displaystyle 5x^2- 6x- 8= 0$.

The only positive integers of 5 are 5 and 1. -8 can be factored as (-1)(8), (-2)(4), (1)(-8), and (2)(-4). Try those.

4. I've worked it out! Thanks everyone! I got (5x+4)(2x-4) so 5x=-4 or 2x=4 so x= -4/5 or x=4/2

5. Originally Posted by BabyMilo
Solve Algebraically - These Simultaneous equations.
...
10x^2 - 12x -16=0

Can anyone solve this without using the quadratic equation?
The quadratic formula method would be easier & faster.

However
Look at the factors of 10: 1 & 10, 2 & 5, 2.5 & 4, 3 & 3 1/3 etc
then the factors of -12: 1 & 12, 2 & 6, 3 & 4, etc
then the factors of -16: 1 & 16, 2 & 8,

If you are logical & systematic, you can reduce the total amount of trials required to arrive at a workable solution.

With a little work you'll end with (2x-4)*(5x+4)

Then solve for each x within the parentheses:
such as:
5x+4 = 0
5x = -4
x = -4/5
Like that.

Why not use the quadratic formula?

6. It's a non-calculator exam paper!